Sum of Two Periodic Orthogonal Functions

[Dschumanji]

Homework Statement

This problem is not from a textbook, it is something I have been thinking about after watching some lectures on Fourier series, the Fourier transform, and the Laplace transform. Suppose you have a real valued periodic function f with fundamental period R and a real valued periodic function g with fundamental period R/m where m is a positive integer not equal to 1. Also suppose that the integral of the product of f and g over the interval [0, R] is equal to 0 (f and g are orthogonal). Can the fundamental period of f+g be less than R?

Homework Equations

The Attempt at a Solution

If m is equal to 1 or you do not assume orthogonality, then it is simple to produce examples where f+g has a fundamental period less than R. I can't seem to find an example of f and g being orthogonal and their sum having a fundamental period less than R. All of the examples I can think of must have a fundamental period of R. This is leading me to believe that f+g cannot have a fundamental period that is less than R if f and g are orthogonal. All of my attempts to prove that the fundamental period must be R have failed. I feel that there is maybe not enough information given to form a proof.

 

[Dr. Courtney]

I tend to take an experimental approach to problems like this and search for a counter example.

 

[STEMucator]

Suppose you have a real valued periodic function f with fundamental period R and a real valued periodic function g with fundamental period R/m where m is a positive integer not equal to 1. Also suppose that the integral of the product of f and g over the interval [0, R] is equal to 0 (f and g are orthogonal). Can the fundamental period of f+g be less than R?

Let $T$ be the period of $f$ instead of $R$. Then:

$$f: \mathbb{R} \rightarrow \mathbb{R} \space | \space f(x + T) = f(x), \space \forall x \in \mathbb{R}$$
$$g: \mathbb{R} \rightarrow \mathbb{R} \space | \space g\left(x + \frac{T}{m}\right) = g(x), \space \forall x \in \mathbb{R}, \space m \in \mathbb{N}, \space m > 0, \space m \neq 1$$

Usually $T$ is used to denote period. The period of $f + g$ is going to be given by:

$$\max\{T, \frac{T}{m} \}$$

Thinking about it a bit, suppose $m = 2$. Then $f$ has period $T$ and $g$ has period $\frac{T}{2}$. These two functions will intersect each other at $x = T$ regardless of $m$. If we took $m = 3, 4, ...$ it wouldn't make a difference.

Taking a more practical example, let $f(x) = \text{sin}(x)$ and $g(x) = \text{sin}(2x)$. Then $f(x)$ has period $2 \pi$ and $g(x)$ has period $\pi$ (where $T = 2 \pi$ and $m = 2$). The period of the sum $f(x) + g(x)$ is going to be $2 \pi$.

The functions will be orthogonal over the interval $[0, 2 \pi]$:

$$\left< f, g \right> = \int_0^{2 \pi} f * g \space dx = 0$$

So the period of the sum is the larger of the two periods.

 

[Dschumanji]

I tend to take an experimental approach to problems like this and search for a counter example.

I have been trying to find examples of orthogonal functions whose sum has a fundamental period less than R. I have had no luck so far.

 

[Dschumanji]

Let $T$ be the period of $f$ instead of $R$. Then:

$$f: \mathbb{R} \rightarrow \mathbb{R} \space | \space f(x + T) = f(x), \space \forall x \in \mathbb{R}$$
$$g: \mathbb{R} \rightarrow \mathbb{R} \space | \space g\left(x + \frac{T}{m}\right) = g(x), \space \forall x \in \mathbb{R}, \space m \in \mathbb{N}, \space m > 0, \space m \neq 1$$

Usually $T$ is used to denote period. The period of $f + g$ is going to be given by:

$$\max\{T, \frac{T}{m} \}$$

Thinking about it a bit, suppose $m = 2$. Then $f$ has period $T$ and $g$ has period $\frac{T}{2}$. These two functions will intersect each other at $x = T$ regardless of $m$. If we took $m = 3, 4, ...$ it wouldn't make a difference.

Taking a more practical example, let $f(x) = \text{sin}(x)$ and $g(x) = \text{sin}(2x)$. Then $f(x)$ has period $2 \pi$ and $g(x)$ has period $\pi$ (where $T = 2 \pi$ and $m = 2$). The period of the sum $f(x) + g(x)$ is going to be $2 \pi$.

The functions will be orthogonal over the interval $[0, 2 \pi]$:

$$\left< f, g \right> = \int_0^{2 \pi} f * g \space dx = 0$$

So the period of the sum is the larger of the two periods.

It is easy to prove that the period is T, but it does not prove it is the fundamental period.

 

[STEMucator]

It is easy to prove that the period is T, but it does not prove it is the fundamental period.

Period and fundamental period are the same concept. Period is defined as the amount of time it takes for one cycle of an event to occur before it repeats itself. If you superimpose two waves for example, the period of the superimposed wave will be the larger of the two waves.

 

[verty]

Dschumanji: try to find a shortcut, there is a shortcut. This question is actually almost trivial.

 

[Dschumanji]

Period and fundamental period are the same concept. Period is defined as the amount of time it takes for one cycle of an event to occur before it repeats itself. If you superimpose two waves for example, the period of the superimposed wave will be the larger of the two waves.

They are not the same concept. If there exists a value T such that f(x+T)=f(x) for all x, then f has a period of T. If P is the smallest value such that f(x+P)=f(x) for all x, then P is the fundamental period of f. For example, cos(x) has a period of 4*pi, but it's fundamental period is 2*pi. The function f defined as being 1 when x is rational and 0 when x is irrational has an infinite number of periods, but no fundamental period.

There are examples of functions whose sum does not have a period that is the larger of the two periods. For example if f(x)=cos(x)-cos((2/3)x) and g(x)=cos((2/3)x), then their sum is cos(x). The fundamental period of f is 6*pi and the fundamental period of g is 3*pi, but the sum of f and g has a fundamental period that is 2*pi. However, this is is an example where f and g are not orthogonal.

 

[Dschumanji]

Dschumanji: try to find a shortcut, there is a shortcut. This question is actually almost trivial.

Any hint as to what that shortcut may be?

 

[verty]

Any hint as to what that shortcut may be?

You haven't actually showed us any work in progress yet. You've asserted that you've tried stuff but we haven't seen it. I don't think I want to say more until you show us what you have tried.

 

[verty]

This is leading me to believe that f+g cannot have a fundamental period that is less than R if f and g are orthogonal. All of my attempts to prove that the fundamental period must be R have failed. I feel that there is maybe not enough information given to form a proof.

This suggests you have certainly tried a bunch of stuff. This is what I want to see.

 

[Dschumanji]

You haven't actually showed us any work in progress yet. You've asserted that you've tried stuff but we haven't seen it. I don't think I want to say more until you show us what you have tried.

Fair enough. I have been trying to prove that sum of f and g must be equal to R if f and g are orthogonal. My method is to use proof by contradiction. I start by assuming that f+g has a fundamental period T that is less than R. Since T is the fundamental period of f+g and f+g has a period R, the fundamental period of f+g must satisfy the equation nT=R for some non-negative integer that is not equal to 1. If I suppose that n=m, then (f+g)(x+(1/n)R)=f(x+(1/n)R)+g(x+(1/n)R)=f(x+(1/n)R)+g(x+(1/m)R)=f(x+(1/n)R)+g(x). However, (f+g)(x+(1/n)R)=f(x)+g(x). Therefore you can conclude that (f)(x+(1/n)R)=f(x). This says that f has a period of (1/n)R. This is impossible since the fundamental period of f is R. I don't know where else to go from here. I tried to playing around with the fact that f and g are orthogonal, but have no idea what to do with it. I can show that the integral of (f+g)^2 over [0, R] is equal to the integral of f^2 over [0, R] plus the integral of g^2 over [0, R].

 

[verty]

Fair enough. I have been trying to prove that sum of f and g must be equal to R if f and g are orthogonal. My method is to use proof by contradiction. I start by assuming that f+g has a fundamental period T that is less than R. Since T is the fundamental period of f+g and f+g has a period R, the fundamental period of f+g must satisfy the equation nT=R for some non-negative integer that is not equal to 1. If I suppose that n=m, then (f+g)(x+(1/n)R)=f(x+(1/n)R)+g(x+(1/n)R)=f(x+(1/n)R)+g(x+(1/m)R)=f(x+(1/n)R)+g(x). However, (f+g)(x+(1/n)R)=f(x)+g(x). Therefore you can conclude that (f)(x+(1/n)R)=f(x). This says that f has a period of (1/n)R. This is impossible since the fundamental period of f is R. I don't know where else to go from here. I tried to playing around with the fact that f and g are orthogonal, but have no idea what to do with it. I can show that the integral of (f+g)^2 over [0, R] is equal to the integral of f^2 over [0, R] plus the integral of g^2 over [0, R].

I think this was a pretty good attempt so I will hint at my shortcut.

Hint: shift(f+g) = shift(f) + shift(g)

 

[Dschumanji]

I think this was a pretty good attempt so I will hint at my shortcut.

Hint: shift(f+g) = shift(f) + shift(g)

Hmmm, I will see what I can do with this. Thank you! When you say something like shift(f), do you mean like f(x+A) or f(x)+A for some value A?

 

[verty]

Hmmm, I will see what I can do with this. Thank you! When you say something like shift(f), do you mean like f(x+A) or f(x)+A for some value A?

I mean the first one, a shift to the left by whatever amount.

 

[Dschumanji]

I mean the first one, a shift to the left by whatever amount.

I made a little bit of progress. I was able to prove that if the fundamental period of f+g is less than R, then it must be of the form R/n where n is an integer such that n>1, m does not divide n, and n does not divide m. To do this, I wrote f(x) as f(x) = (f+g)(x) - g(x) and replaced x with x+R/m or x+R/n (the former for assuming m|n and the latter for assuming n|m). This leads to contradictions with the fundamental period of f.

You said this problem is almost trivial but I just don't see it. It would be trivial to prove f+g has period R, but that doesn't prove it is the fundamental period of f+g.

 

[verty]

Ok, I'll show you the shortcut, I know this isn't a homework problem. I think you'll agree that it's quite amazing this question has such a simple answer when it looks so daunting at the beginning.

Let P be a period of f+g. Let "shift" denote a left shift by P.
Then f+g = shift(f+g) = shift(f) + shift(g) = shift(f) + g
Therefore f = shift(f).
But that is only true when P $\geq$ R. Therefore P $\geq$ R, QED.

 

[Dschumanji]

Ok, I'll show you the shortcut, I know this isn't a homework problem. I think you'll agree that it's quite amazing this question has such a simple answer when it looks so daunting at the beginning.

Let P be a period of f+g. Let "shift" denote a left shift by P.
Then f+g = shift(f+g) = shift(f) + shift(g) = shift(f) + g
Therefore f = shift(f).
But that is only true when P $\geq$ R. Therefore P $\geq$ R, QED.

Isn't the following a counter example to your claim, though?

"There are examples of functions whose sum does not have a period that is the larger of the two periods. For example if f(x)=cos(x)-cos((2/3)x) and g(x)=cos((2/3)x), then their sum is cos(x). The fundamental period of f is 6*pi and the fundamental period of g is 3*pi, but the sum of f and g has a fundamental period that is 2*pi. However, this is is an example where f and g are not orthogonal."

 

[verty]

Isn't the following a counter example to your claim, though?

"There are examples of functions whose sum does not have a period that is the larger of the two periods. For example if f(x)=cos(x)-cos((2/3)x) and g(x)=cos((2/3)x), then their sum is cos(x). The fundamental period of f is 6*pi and the fundamental period of g is 3*pi, but the sum of f and g has a fundamental period that is 2*pi. However, this is is an example where f and g are not orthogonal."

Darn, you're right.

Sorry, I thought there was a simple fix but I was wrong about that. The problem is that one needs the lemma: if P is a period of f+g, then P is a period of g. Otherwise one can't say that g = shift(g).

And looking at your last post, to prove this lemma requires showing that R is the fundamental period of f+g, so my method is in fact completely useless.

 

[verty]

Let f have a fundamental period of $12\pi$. We can break it up into six pieces, each with a size of $2\pi$: abcdef.

Let the following be true:
a = sin(2x),
b = sin(3x) + sin(x),
c = sin(2x) - sin(x),
d = sin(3x),
e = sin(2x) + sin(x),
f = sin(3x) - sin(x).

Similarly, let g have a fundamental period of $4\pi$ and be made up of two pieces of size $2\pi$:
a = sin(3x),
b = sin(2x).

Now it should happen that f and g are orthogonal with fundamental periods of $12\pi$ and $4\pi$ and that f+g has a fundamental period of $6\pi$.

 

[Dschumanji]

Let f have a fundamental period of $12\pi$. We can break it up into six pieces, each with a size of $2\pi$: abcdef.

Let the following be true:
a = sin(2x),
b = sin(3x) + sin(x),
c = sin(2x) - sin(x),
d = sin(3x),
e = sin(2x) + sin(x),
f = sin(3x) - sin(x).

Similarly, let g have a fundamental period of $4\pi$ and be made up of two pieces of size $2\pi$:
a = sin(3x),
b = sin(2x).

Now it should happen that f and g are orthogonal with fundamental periods of $12\pi$ and $4\pi$ and that f+g has a fundamental period of $6\pi$.

Verty, you're a genius! I've spent days on this problem and you solved it in one night. I clearly need to get better at coming up with counter examples! Thank you so much for your help!

 

[verty]

Verty, you're a genius! I've spent days on this problem and you solved it in one night. I clearly need to get better at coming up with counter examples! Thank you so much for your help!

Thanks, it was certainly perplexing why was so impervious to being solved. There was an olympiad question once, I happened to look at the paper, the question was extremely tough but the only way to solve it was using linear algebra techniques, producing a system of equations and listing constraints between the variables.

I thought of that and decided to carve up the function like that into 8 letters and have a system like
a b c d e f j k
[ 1 0 0 -1 0 0 1 -1 ], etc.

And it quickly became obvious that there should be infinitely many solutions. Then it was just a matter of picking the easiest functions for each variable and to do a bit of tweaking.

I'm glad to have helped and thank you very much for the interesting question.

 

[Dschumanji]

Thanks, it was certainly perplexing why was so impervious to being solved. There was an olympiad question once, I happened to look at the paper, the question was extremely tough but the only way to solve it was using linear algebra techniques, producing a system of equations and listing constraints between the variables.

I thought of that and decided to carve up the function like that into 8 letters and have a system like
a b c d e f j k
[ 1 0 0 -1 0 0 1 -1 ], etc.

And it quickly became obvious that there should be infinitely many solutions. Then it was just a matter of picking the easiest functions for each variable and to do a bit of tweaking.

I'm glad to have helped and thank you very much for the interesting question.

That is an interesting approach! I was wondering how you came up with the counter example.